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CHAPTER 1.1 Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az, find: a) a unit vector in the direction of −M + 2N −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus (26, 10, 4) a= = (0.92, 0.36, 0.14) |(26, 10, 4)| b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)|= 48.6 c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2 Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7, −2, 1): a) Specify the vector A extending from the origin to the point A A = (4, 3, 2) = 4ax + 3ay + 2az b) Give a unit vector extending from the origin to the midpoint of line AB The vector from the origin to the midpoint is given by M = (1/2)(A + B) = (1/2)(4 − 2, + 0, + 5) = (1, 1.5, 3.5) The unit vector will be m= (1, 1.5, 3.5) |(1, 1.5, 3.5)| = (0.25, 0.38, 0.89) c) Calculate the length of the perimeter of triangle ABC: Begin with AB = (−6, −3, 3), BC = (9, −2, −4), CA = (3, −5, −1) Then |AB|+ |BC|+ |CA|= 7.35 + 10.05 + 5.91 = 23.32 1.3 The vector from the origin to the point A is given as (6,— 2, −4), and the unit vector directed from the origin toward point B is (2, − 2, 1)/3 If points A and B are ten units apart, find the coordinates of point B With A = (6, −2, −4)2 and B = 31 B(2, −2, 1), we use the fact that |B − A|= 10, or (6 B)a (2 B)a (4 B)a x y z | = 10 | −3 − −3 − +3 Expanding, obtain 36 − 8B + B2 + − B + B2 + 16 + B + B2 = 100 9 or B2 − 8B − 44 = Thus B = 8± √64−176 = 11.75 (taking positive option) and so 2 B = (11.75)a x — (11.75)a y + (11.75)a z = 7.83a — 7.83a + 3.92a 3 x y z 1.4 given points A(8, −5, 4) and B(−2, 3, 2), find: a) the distance from A to B |B − A|= |(−10, 8, −2)|= 12.96 b) a unit vector directed from A towards B This is found through a AB = B−A |B − A| ( 0.77, 0.62, 0.15) = − − c) a unit vector directed from the origin to the midpoint of the line AB (A + B)/2 (3, −1, 3) a0 = = = (0.69, −0.23, 0.69) M √ |(A + B)/2| 19 d) the coordinates of the point on the line connecting A to B at which the line intersects the plane=z Note that the midpoint, (3, −1, 3), as determined from part c happens to have z coordinate of This is the point we are looking for 1.5 A vector field is specified as G = 24xyax + 12(x2 + 2)ay + 18z2az Given two points, P (1, 2, −1) and Q(−2, 1, 3), find: a) G at P : G(1, 2, −1) = (48, 36, 18) b) a unit vector in the direction of G at Q: G(−2, 1, 3) = (−48, 72, 162), so a G = (−48, 72, 162) |(−48, 72, 162)| = (−0.26, 0.39, 0.88) c) a unit vector directed from Q toward P : P−Q (3, −1, 4) a = = = (0.59, 0.20, −0.78) QP √ |P − Q| 26 d) the equation of the surface on which |G| = 60: We write 60 = |(24xy, 12(x2 + 2), 18z2)|, or 10 = |(4xy, 2x2 + 4, 3z2)|, so the equation is 100 = 16x2y2 + 4x4 + 16x2 + 16 + 9z4 1.6 For the G field in Problem 1.5, make sketches of Gx, Gy, Gz and |G| along the line y = 1, z = 1, for ≤ x ≤ We find √ G(x, 1, 1) = (24x, 12x2 + 24, 18), from which Gx = 24x, Gy = 12x2 + 24, Gz = 18, and |G|= 4x + 32x + 25 Plots are shown below 1.7 Given the vector field E = 4zy2 cos 2xax + 2zy sin 2xay + y2 sin 2xaz for the region x| ,| y| , |and z |less | than 2, find: a) the surfaces on which Ey = With Ey = 2zy sin 2x = 0, the surfaces are 1) the plane z = 0, with |x| < 2, |y| < 2; 2) the plane y = 0, with |x| < 2, |z| < 2; 3) the plane x = 0, with |y| < 2, |z| < 2; 4) the plane x = π/2, with |y| < 2, |z| < b) the region in which Ey = Ez: This occurs when 2zy sin 2x = y2 sin 2x, or on the plane 2z = y, with |x| < 2, |y| < 2, |z| < c) the region in which E = 0: We would have Ex = Ey = Ez = 0, or zy2 cos 2x = zy sin 2x = y2 sin 2x = This condition is met on the plane y = 0, with |x| < 2, |z| < 1.8 Two vector fields are F = −10ax + 20x(y − 1)ay and G = 2x2yax − 4ay + zaz For the point P (2, 3, −4), find: a) |F|: F at (2, 3, −4) = (−10, 80, 0), so |F|= 80.6 b) |G|: G at (2, 3, −4) = (24, −4, −4), so |G|= 24.7 c) a unit vector in the direction of F − G: F − G = (−10, 80, 0) − (24, −4, −4) = (−34, 84, 4) So a= F−G = (−34, 84, 4) |F − G| = (−0.37, 0.92, 0.04) 90.7 d) a unit vector in the direction of F + G: F + G = (−10, 80, 0) + (24, −4, −4) = (14, 76, −4) So a= F+G = (14, 76, −4) |F + G| 77.4 = (0.18, 0.98, −0.05) 1.9 A field is given as G= 25 (x2 + y2) (xax + yay) Find: a) a unit vector in the direction of G at P (3, 4, −2): Have Gp = 25/(9 + 16) × (3, 4, 0) = 3ax + 4ay, and |Gp | = Thus aG = (0.6, 0.8, 0) b) the angle between G and ax at P : The angle is found through aG · ax = cos θ So cos θ = (0.6, 0.8, 0) · (1, 0, 0) = 0.6 Thus θ = 53◦ c) the value of the following double integral on the plane y = 7: ∫ 4∫ ∫ 4∫ 25 x2 + y2 G · a y dzdx (xax + yay) · aydzdx = ∫ 4∫ = 350 × 1tan−1 ∫ 25 x2 + 49 × dzdx = 350 dx x2 + 49 — = 26 1.10 Use the definition of the dot product to find the interior angles at A and B of the triangle defined by the three points A(1, 3, −2), B(−2, 4, 5), and C(0, −2, 1): a) Use RAB = (− , 7) and R AC = (−1, −5, 3) to form RAB · RAC = |RAB ||RAC | cos θA Obtain √3, 1√ + + 21 = 59 35 cos θA Solve to find θA = 65.3◦ b) Use RBA = (3√ , −1√ , −7) and R B C = (2, −6, −4) to form RBA · RBC = |RBA ||RBC | cos θB Obtain + + 28 = 59 56 cos θB Solve to find θB = 45.9◦ 1.11 Given the points M(0.1, −0.2, −0.1), N(−0.2, 0.1, 0.3), and P (0.4, 0, 0.1), find: a) the vector RMN : RMN = (−0.2, 0.1, 0.3) − (0.1, −0.2, −0.1) = (−0.3, 0.3, 0.4) b) the dot product RMN · RMP : RMP = (0.4, 0, 0.1) − (0.1, −0.2, −0.1) = (0.3, 0.2, 0.2) RMN · RMP = (−0.3, 0.3, 0.4) · (0.3, 0.2, 0.2) = −0.09 + 0.06 + 0.08 = 0.05 c) the scalar projection of RMN on RMP : RMN · aRMP = (−0.3, 0.3, 0.4) · √ 0.05 =√ 0.17 = 0.12 0.09 + 0.04 + 0.04 (0.3, 0.2, 0.2) d) the angle between R MN and RMP : R θ M = cos−1 ·R MP |RMN MN ||RMP | 0.05 = cos−1 √0.34√ 0.17 = 78◦ 1.12 Given points A(10, 12, −6), B(16, 8, −2), C(8, 1, −4), and D(−2, −5, 8), determine: a) the vector projection of RAB + RBC on RAD: RAB + RBC = RAC = (8, 1, 4) − (10, 12, −6) = (−2, −11, 10) Then RAD = (−2, −5, 8) − (10, 12, −6) = (−12, −17, 14) So the projection will be: (RAC · aRAD )aRAD = (−2, −11, 10) · (−12, −17, √ 14) 629 (−12,√−17, 14) = (−6.7, −9.5, 7.8) 629 b) the vector projection of RAB + RBC on RDC: RDC = (8, − 1, 4) − (− 2, − 5, 8) = (10, 6, − 4) The projection is: (RAC · aRDC )aRDC = (−2, −11, 10) · (10, √ 6, −4) 152 (10,√6, −4) 152 = (−8.3, −5.0, 3.3) c) the angle between RDA and RDC: Use RDA = − RAD = (12, 17, −14) and RDC = (10, 6, − 4) The angle is found through the dot product of the associated unit vectors, or: θD = cos−1 (aRDA · aRDC ) = cos−1 (12, 17, −14) · (10, 6, −4) √ √ 629 152 = 26◦ 1.13 a) Find the vector component of F = (10, −6, 5) that is parallel to G = (0.1, 0.2, 0.3): F·G (10, −6, 5) · (0.1, 0.2, 0.3) F = G= (0.1, 0.2, 0.3) = (0.93, 1.86, 2.79) ||G |G|2 0.01 + 0.04 + 0.09 b) Find the vector component of F that is perpendicular to G: FpG = F − F||G = (10, −6, 5) − (0.93, 1.86, 2.79) = (9.07, −7.86, 2.21) c) Find the vector component of G that is perpendicular to F: G·F 1.3 = G− F = (0.1, 0.2, 0.3) − G = G−G (10, −6, 5) = (0.02, 0.25, 0.26) pF ||F |F| 100 + 36 + 25 √ 1.14 The√four vertices of a regular tetrahedron are located at O(0, 0, 0), A(0, 1, 0), B(0.5 3, 0.5, 0), and √ C( 3/6, 0.5, 2/3) a) Find a unit vector perpendicular (outward) to the face ABC: First find √ √ , √ 0) − (0.5 3, 0.5, RBA × RBC = [(0, 1, √ √0)] × [(, 3/6, 0.5, 2/3) − (0.5 3, 0.5, 0)] = (−0.5 3, 0.5, 0) × ( 3/3, 0, /3) = (0.41, 0.71, 0.29) — The required unit vector will then be: RBA × RBC |RBA × RBC| = (0.47, 0.82, 0.33) b) Find the area of the face ABC: Area R × R | = 0.43 BC = | BA 1.15 Three vectors extending from the origin are given as r1 = (7, 3, −2), r2 = (−2, 7, −3), and r3 = (0, 2, 3) Find: a) a unit vector perpendicular to both r1 and r2: r1 × r2 (5, 25, 55) = = (0.08, 0.41, 0.91) ap12 = |r1 × r2| 60.6 b) a unit vector perpendicular to the vectors r1 − r2 and r2 − r3: r1 − r2 = (9, −4, 1) and r2 − r3 = (−2, 5, −6) So r1 − r2 × r2 − r3 = (19, 52, 32) Then (19, 52, 32) (19, 52, 32) = = (0.30, 0.81, 0.50) ap = 63.95 |(19, 52, 32)| c) the area of the triangle defined by r1 and r2: Area = |r × r |= 30.3 d) the area of the triangle defined by the heads of r1, r2, and r3: 1 Area = |(r2 − r1) × (r2 − r )|= |(−9, 4, −1) × (−2, 5, −6)|= 32.0 2 1.16 Describe the surfaces defined by the equations: a) r · ax = 2, where r = (x, y, z): This will be the plane x = , b) |r × ax | = 2: r × ax = (0, z, −y), and |r × ax | = z2 + y2 = This is the equation of a cylinder, centered on the x axis, and of radius 1.17 Point A(−4, 2, 5) and the two vectors, RAM = (20, 18, −10) and RAN = (−10, 8, 15), define a triangle a) Find a unit vector perpendicular to the triangle: Use ap = RAM × RAN |RAM × RAN | = (350, −200, 340) 527.35 = (0.664, − 0.379, 0.645) The vector in the opposite direction to this one is also a valid answer b) Find a unit vector in the plane of the triangle and perpendicular to RAN : √ 8, 15) = (−0.507, 0.406, 0.761) a AN = (−10, 389 Then apAN = ap × aAN = (0.664, −0.379, 0.645) × (−0.507, 0.406, 0.761) = (−0.550, −0.832, 0.077) The vector in the opposite direction to this one is also a valid answer c) Find a unit vector in the plane of the triangle that bisects the interior angle at A: A non-unit vector in the required direction is (1/2)(aAM + aAN ), where aAM = (20, 18, −10) = (0.697, 0.627, −0.348) |(20, 18, −10)| 1.17c (continued) Now (aAM + aAN ) = [(0.697, 0.627, −0.348) + (−0.507, 0.406, 0.761)] = (0.095, 0.516, 0.207) Finally, (0.095, 0.516, 0.207) abis = |(0.095, 0.516, 0.207)| = (0.168, 0.915, 0.367) 1.18 Given points A(ρ = 5,φ = 70 ◦,z = −3) and B(ρ = 2,φ = −30◦,z = 1), find: a) unit vector in cartesian coordinates at A toward B: A(5 cos 70◦, sin 70◦, − 3) = A(1.71, 4.70, − 3), In the same manner, B(1.73, − 1, 1) So RAB = (1.73, − 1, 1) − (1.71, 4.70, − 3) = (0.02, − 5.70, 4) and therefore (0.02, −5.70, 4) a AB = = (0.003, −0.82, 0.57) |(0.02, −5.70, 4)| b) a vector in cylindrical coordinates at A directed toward B: aAB · aρ = 0.003 cos 70◦ − 0.82 sin 70◦ = −0.77 aAB · aφ = −0.003 sin 70◦ − 0.82 cos 70◦ = −0.28 Thus aAB = −0.77aρ − 0.28aφ + 0.57az c) a unit vector in cylindrical coordinates at B directed toward A: Use aBA = (−0, 003, 0.82, −0.57) Then aBA · aρ = −0.003 cos(−30◦) + 0.82 sin(−30◦) = −0.43, and aBA · aφ = 0.003 sin(−30◦) + 0.82 cos(−30◦) = 0.71 Finally, aBA = −0.43aρ + 0.71aφ − 0.57az 1.19 a) Express the field D = (x2 + y2)−1(xax + yay) in cylindrical components and cylindrical variables: Have x = ρ cos φ, y = ρ sin φ, and x2 + y2 = ρ2 Therefore D = (cos φa + sin φa ) x y ρ Then Dρ = D · aρ = ρ cos φ(ax · aρ) + sin φ(ay · aρ) = ρ 2 cos φ + sin φ = ρ and Dφ = D · aφ = ρ cos φ(ax · aφ) + sin φ(ay · aφ) = Therefore D= aρ ρ ρ [cos φ(− sin φ) + sin φ cos φ] = 1.19b Evaluate D at the point where ρ = 2, φ = 0.2π, and z = 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D = 0.5aρ To express this in cartesian, we use D = 0.5(aρ · ax)ax + 0.5(aρ · ay)ay = 0.5 cos 36◦ax + 0.5 sin 36◦ay = 0.41ax + 0.29ay 1.20 Express in cartesian components: a) the vector at A(ρ = 4,φ = 40 ◦ ,z = −2) that extends to B(ρ = 5,φ = −110◦,z = 2): We have A(4 cos 40◦, sin 40◦, −2) = A(3.06, 2.57, −2), and B(5 cos(−110◦), sin(−110◦), 2) = B(−1.71, −4.70, 2) in cartesian Thus RAB = (−4.77, −7.30, 4) b) a unit vector at B directed toward A: Have RBA = (4.77, 7.30, −4), and so a BA = (4.77, 7.30, −4) = (0.50, 0.76, −0.42) |(4.77, 7.30, −4)| Have rB = (−1.71, −4.70, 2), and so −rB = c) a unit vector at B directed toward the origin: (1.71, 4.70, −2) Thus a= (1.71, 4.70, −2) |(1.71, 4.70, −2)| = (0.32, 0.87, −0.37) 1.21 Express in cylindrical components: a) the vector from C(3, 2, −7) to D(−1, −4, 2): C(3, 2, −7) → C(ρ = 3.61,φ = 33.7◦,z = −7) and D(−1, −4, 2) → D(ρ = 4.12,φ = −104.0◦,z = 2) Now RCD = (−4, −6, 9) and Rρ = RCD · aρ = −4 cos(33.7) − sin(33.7) = −6.66 Then Rφ = RCD · aφ = sin(33.7) − cos(33.7) = −2.77 So RCD = −6.66aρ − 2.77aφ + 9az b) a unit vector at D directed toward C: RCD = (4, 6, −9) and Rρ = RDC · aρ = cos(−104.0) + sin(−104.0) = −6.79 Then Rφ = RDC · aφ = 4[− sin(−104.0)] + cos(−104.0) = 2.43 So RDC = −6.79aρ + 2.43aφ − 9az Thus aDC = −0.59aρ + 0.21aφ − 0.78az c) a unit vector at D directed toward the origin: Start with rD = ( −1, −4, 2), and so the vector toward the origin will be − rD =(1, 4, −2) Thus in cartesian the unit vector is a = (0.22, 0.87, − 0.44) Convert to cylindrical: aρ = (0.22, 0.87, −0.44) · aρ = 0.22 cos(−104.0) + 0.87 sin(−104.0) = −0.90, and aφ = (0.22, 0.87, −0.44) · aφ = 0.22[− sin(−104.0)] + 0.87 cos(−104.0) = 0, so that finally, a = −0.90aρ − 0.44az 1.22 A field is given in cylindrical coordinates as F= 40 ρ2 + + 3(cos φ + sin φ) aρ + 3(cos φ − sin φ)aφ − 2az where the magnitude of F is found to be: √ |F|= 1600 F·F= 240 (ρ2 + 1)2 + ρ2 + 1/2 (cos φ + sin φ) + 22 Sketch |F|: a) vs φ with ρ = 3: in this case the above simplifies to |F(ρ = 3)|= |F a| = [38 + 24(cos φ + sin φ)] 1/2 b) vs ρ with φ = 0, in which: |F(φ = 0)|= |F b| = 1600 (ρ2 + 1)2 240 + c) vs ρ with φ = 45◦, in which |F(φ = 45 ◦ )|= |F c | = 1600 (ρ2 + 1)2 ρ2 + 1/2 + 22 √ 240 + ρ2 + 1/2 + 22 1.23 The surfaces ρ = 3, ρ = 5, φ = 100◦, φ = 130◦, z = 3, and z = 4.5 define a closed surface a) Find the enclosed volume: ∫ 4.5 ∫ 130◦ ∫ Vol = 100 ◦ 3 ρ dρ dφ dz = 6.28 NOTE: The limits on the φ integration must be converted to radians (as was done here, but not shown) b) Find the total area of the enclosing surface: ∫ ∫ 130◦ ∫ Area = ρ dρ dφ + ◦ + ∫ 100 4.5 ∫ 130◦ 100◦ ∫ 4.5 ∫ 130◦ ◦ 4.5 ∫ 5 dφdz + 3 d φd z 100 dρ dz = 20.7 c) Find the total length of the twelve edges of the surfaces: Length = × 1.5 + × + × 30◦ 360◦ × 2π × + 30◦ 360◦ × 2π × = 22.4 d) Find the length of the longest straight line that lies entirely within the volume: This will be between the points A(ρ = 3, φ = 100◦, z = 3) and B(ρ = 5, φ = 130◦, z = 4.5) Performing point transformations to cartesian coordinates, these become A(x = −0.52, y = 2.95, z = 3) and B(x = −3.21, y = 3.83, z = 4.5) Taking A and B as vectors directed from the origin, the requested length is Length = |B − A|= |(−2.69, 0.88, 1.5)|= 3.21 1.24 At point P (−3, 4, 5), express the vector that extends from P to Q(2, 0, −1) in: a) rectangular coordinates Then |R PQ |= RPQ = Q − P = 5ax − 4ay − 6az √ 25 + 16 + 36 = 8.8 b) cylindrical coordinates At P , ρ = 5, φ = tan−1(4/ − 3) = −53.1◦, and z = Now, RPQ · aρ = (5ax − 4ay − 6az) · aρ = cos φ − sin φ = 6.20 RPQ · aφ = (5ax − 4ay − 6az) · aφ = −5 sin φ − cos φ = 1.60 Thus RPQ = 6.20aρ + 1.60aφ − 6az √ and |R PQ |= 6.202 + 1.602 + 62 = 8.8 √ √ c) spherical coordinates At P , r = + 16 + 25 = 50 = 7.07, θ = cos−1(5/7.07) = 45◦, and φ = tan−1(4/ − 3) = −53.1◦ RPQ · ar = (5ax − 4ay − 6az) · ar = sin θ cos φ − sin θ sin φ − cos θ = 0.14 RPQ · aθ = (5ax − 4ay − 6az) · aθ = cos θ cos φ − cos θ sin φ − (−6) sin θ = 8.62 RPQ · aφ = (5ax − 4ay − 6az) · aφ = −5 sin φ − cos φ = 1.60 10 13.39 (continued) l 5l − V + = 0.44 V0 n2 (note error in problem statement) a) Write, in terms of the appropriate indices, an expression for the minimum possible wave angle, θ1, that a guided mode may have: The wave angle must be equal to or greater than the critical angle of total reflection at both interfaces The minimum wave angle is thus determined by the greater of the two critical angles Since n3 > n2, we find θmin = θc,13 = sin−1(n3/n1) b) Write an expression for the maximum phase velocity a guided mode may have in this structure, using given or known parameters: We have vp,max = ω/βmin, where βmin = n1k0 sin θ1,min = n1k0n3/n1 = n3k0 Thus vp,max = ω/(n3k0) = c/n3 14.20 A step index optical fiber is known to be single mode at wavelengths λ > 1.2 µm Another fiber is to be fabricated from the same materials, but is to be single mode at wavelengths λ > 0.63 µm By what percentage must the core radius of the new fiber differ from the old one, and should it be larger or smaller? We use the cutoff condition, given by (80): , 2πa n2 − n 2 λ> 2.405 255 14.20 (continued) With λ reduced, the core radius, a, must also be reduced by the same fraction Therefore, the percentage reduction required in the core radius will be % = 1.2 − 63 × 1.2 100 = 47.5% 14.21 A short dipole carrying current I0 cos ωt in the az direction is located at the origin in free space a) If β = rad/m, r =2 m, θ =45◦, φ =0, and t =0, give a unit vector in rectangular components that shows the instantaneous direction of E: In spherical coordinates, the components of E are given by (82) and (83): I dη E r = cos θe−j 2πr/λ 2π Eθ = λ r j 2πr + I0 dη 2π λ + + sin θ e −j 2π r/λ j λr r j 2πr3 4π Since we want a unit vector at t = 0, we need only the relative amplitud√ es of the two components, but we need the absolute phases Since θ = 45◦, sin θ = cos θ = 1/ Also, with β = = 2π/λ, it follows that λ = 2π m The above two equations can be simplified by these substitutions, while dropping all amplitude terms that are common to both Obtain Ar = Aθ = j r + r + jr r + e−jr jr e−jr Now with r = m, we obtain Ar = −j e−j = ◦ (1.12)e−j 26.6 e−j 1 j 56.3◦ −j e Aθ = j + − j e−j = (0.90)e 16 The total vector is now A = Ar ar + Aθ aθ We can normalize the vector by first finding the magnitude: 1, (1.12)2 + (0.90)2 = 0.359 Dividing the field vector by this magnitude and converting rad to 114.6◦, we write the normalized vector as |A| = √ A · A∗ = ◦ ◦ A N s = 0.780e −j 141.2 ar + 0.627e −58.3 aθ In real instantaneous form, this becomes jωt AN (t) = Re ANse = 0.780 cos(ωt − 141.2◦)ar + 0.627 cos(ωt − 58.3◦)aθ We evaluate this at t = to find AN (0) = 0.780 cos(141.2◦)ar + 0.627 cos(58.3◦)aθ = −0.608ar + 0.330aθ 256 14.21a (continued) , Dividing by the magnitude, (0.608)2 + (0.330)2 = 0.692, we obtain the unit vector at t = 0: aN (0) = −0.879ar + 0.477aθ We next convert this to cartesian components: aNx = aN (0) · ax = −0.879 sin θ cos φ + 0.477 cos θ cos φ = √ (−0.879 + 0.477) = −0.284 aNy = aN (0) · ay = −0.879 sin θ sin φ + 0.477 cos θ sin φ = since φ = aNz = aN (0) · az = −0.879 cos θ − 0.477 sin θ = √ (−0.879 − 0.477) = −0.959 The final result is then aN (0) = −0.284ax − 0.959az b) What fraction of the total average power is radiated in the belt, 80◦ < θ < 100◦? We use the far-zone phasor fields, (84) and (85), and first find the average power density: I 2d 2η ∗ 2 Pavg = Re[E H ] = sin θ W/m θs φs 8λ r 2 We integrate this over the given belt, an at radius r: ∫ 2π ∫ 100◦ 2 ∫ 100◦ I d η πI d2 η 0 sin2 θ r2 sin θ dθ dφ = sin3 θ dθ Pbelt = 8λ2 r 4λ2 80◦ 80◦ Evaluating the integral, we find πI d η Pbelt = 4λ2 − cos θ sin2 θ + πI d η 100 = (0.344) 4λ2 80 The total power is found by performing the same integral over θ , where < θ < 180◦ Doing this, it is found that πI 02d2η Ptot = (1.333) 4λ2 The fraction of the total power in the belt is then f = 0.344/1.333 = 0.258 14.22 Prepare a curve, r vs θ in polar coordinates, showing the locus in the φ = plane where: a) the radiation field |Eθs |is one-half of its value at r = 104 m, θ = π/2: Assuming the far field approximation, we use (84) to set up the equation: I dη |E θs |= 2λr I dη sin θ = × ⇒ r = × 10 sin θ × 104λ b) the average radiated power density, Pr,av, is one-half of its value at r = 104 m, θ = π/2 To find the average power, we use (84) and (85) in ∗ I 2d2η sin θ = × Pr,av = Re{EθsH φs } = 2 4λ 2r 257 √ I 2d2η 4λ2 (10 8) ⇒ r= × 10 sin θ 14.22 (continued) The polar plots for field (r = × 104 sin θ ) and power (r = below Both are circles 14.23 √ × 104 sin θ ) are shown Two short antennas at the origin in free space carry identical currents of cos ωt A, one in the az direction, one in the ay direction Let λ = 2π m and d = 0.1 m Find Es at the distant point: a) (x = 0,y 1000,z 0):= This point lies along the axial direction of the ay antenna, so its = contribution to the field will be zero This leaves the az antenna, and since=θ 90◦, only the Eθs component will be present (as (82) and (83) show) Since we are in the far zone, (84) applies We use θ = 90◦, d = 0.1, λ = 2π, η = η0 = 120π, and r = 1000 to write: Es = Eθs aθ = j 5(0.1)(120π) −j 1000 I0 dη e aθ sin θ e −j 2π r/λ a θ = j 2λr 4π(1000) = j (1.5 × 10−2)e− j1000 aθ = −j (1.5 × 10−2)e−j1000 az V/m b) (0, 0, 1000): Along the z axis, only the ay antenna will contribute to the field Since the distance is the same, we can apply the part a result, modified such the the field direction is in −ay: Es = −j (1.5 × 10−2)e− j1000 ay V/m c) (1000, 0, 0): Here, both antennas will contribute Applying the results of parts a and b, we find Es = −j (1.5 × 10−2)(ay + az) d) Find E at (1000, 0, 0) at t = 0: This is found through jωt E(t) = Re Ese = (1.5 × 10−2) sin(ωt − 1000)(ay + az) Evaluating at t = 0, we find E(0) = (1.5 × 10−2)[− sin(1000)](ay + az) = −(1.24 × 10−2)(ay + az) V/m e) Find |E| at (1000, 0, 0) at t = 0: Taking the magnitude of the part d result, we find |E| = 1.75 × 10−2 V/m 258 14.24 A short current element has d = 0.03λ Calculate the radiation resistance for each of the following current distributions: a) uniform: In this case, (86) applies directly and we find Rrad = 80π d λ = 80π (.03) = 0.711 β 2 b) linear, I (z) = I0(0.5d − |z |)/0.5d: Here, the average current is 0.5I0, and so the average power drops by a factor of 0.25 The radiation resistance therefore is down to one-fourth the value found in part a, or Rrad = (0.25)(0.711) = 0.178 β c) step, I0 for < |z | < 0.25d and 0.5I0 for 0.25d < |z |< 0.5d: In this case the average current on the wire is 0.75I0 The radiated power (and radiation resistance) are down to a factor of (0.75)2 times their values for a uniform current, and so Rrad = (0.75)2(0.711) = 0.400 β 14.25 A dipole antenna in free space has a linear current distribution If the length is 0.02λ, what value of I0 is required to: a) provide a radiation-field amplitude of 100 mV/m at a distance of one mile, at =θ 90◦: With a linear current distribution, the peak current, I0, occurs at the center of the dipole; current decreases linearly to zero at the two ends The average current is thus I0/2, and we use Eq (84) to write: I0 dη0 I0 (0.02)(120π) |Eθ | = 4λr sin(90◦) = (4)(5280)(12)(0.0254) = 0.1 ⇒ I = 85.4 A b) radiate a total power of watt? We use Pavg = rad I0 R where the radiation resistance is given by Eq (86), and where the factor of 1/4 arises from the average current of I0/2: We obtain Pavg = 10π2I 2(0.02) = ⇒ I0 = 5.03 A 14.26 A monopole antenna in free space, extending vertically over a perfectly conducting plane, has a linear current distribution If the length of the antenna is 0.01λ, what value of I0 is required to a) provide a radiation field amplitude of 100 mV/m at a distance of mi, at =θ 90◦: The image antenna below the plane provides a radiation pattern that is identical to a dipole antenna of length 0.02λ The radiation field is thus given by (84) in free space, where θ= 90◦, and with an additional factor of 1/2 included to account for the linear current distribution: 4r|Eθ | I0dη0 |Eθ | = 2λr ⇒ I0 = (d/λ)η0 4(5289)(12 × 0254)(100 × 10−3) = (.02)(377) = 85.4A b) radiate a total power of 1W: For the monopole over the conducting plane, power is radiated only over the upper half-space This reduces the radiation resistance of the equivalent dipole antenna by a factor of one-half Additionally, the linear current distribution reduces the radiation resistance of a dipole having uniform current by a factor of one-fourth Therefore, Rrad is one-eighth the value obtained from (86), or Rrad = 10π2(d/λ)2 The current magnitude is now √ 1/2 2(1) 2Pav 1/2 = I0 = = 7.1A = √ 2 10π (d/λ) Rrad 10 π(.02) 259 14.27 The radiation field of a certain short vertical current element is Eθs = (20/r) sin θ e−j10πr V/m if it is located at the origin in free space a) Find Eθs at P (r = 100,θ = 90 ◦,φ = 30◦): Substituting these values into the given formula, find Eθs = 20 ◦ −j 10π(100) = 0.2e−j 1000π V/m 100 sin(90 )e b) Find Eθs at P if the vertical element is located at A(0.1, 90◦, 90◦): This places the element on the y axis at y 0.1 = As a result of moving the antenna from the origin to y 0.1, =the change in distance to point P is negligible when considering the change in field amplitude, but is not when considering the change in phase Consider lines drawn from the origin to P and from y = 0.1 to These can the be line considered parallel, difference their lengthsand is ◦), with l P 0.1 sin(30lines from y essentially = 0.1 being shorterand by so thisthe amount Theinconstruction = arguments are similar to those used in the discussion of the electric dipole in Sec 4.7 The electric field is now the result of part a, modified by including a shorter distance, r, in the phase term only We show this as an additional phase factor: Eθs = 0.2e−j 1000π ej 10π(0.1 sin 30 = 0.2e−j 1000π ej 0.5π V/m c) Find Eθs at P if identical elements are located at A(0.1, 90◦, 90◦) and B(0.1, 90◦, 270◦): The original element of part b is still in place, but a new one has been added at y 0.1 Again, =− constructing a line between B and P , we find, using the same arguments as in part b, that the length of this line is approximately 0.1 sin(30◦) longer than the distance from the origin to P The part b result is thus modified to include the contribution from the second element, whose field will add to that of the first: Eθs = 0.2e−j1000π ej0.5π + e−j0.5π = 0.2e−j1000π cos(0.5π) = The two fields are out of phase at P under the approximations we have used 260